Titration Of A Weak Base With A Strong Acid - Chemistry LibreTexts

Solution

First, calculate the number of moles of base (analyte) present initially.

\[0.090 \; L \; base \; \times \dfrac{0.6 \; mol \; base}{L \; base \; solution} = 0.054 \; mol \; base \nonumber \]

a) An ICE table helps determine the molarity of OH-.

\(NH_3\)	\(H_2O\)	\(\rightleftharpoons\)	\(NH_4^+\)	\(OH^-\)
Initial	0.6 M	-	0	0
Change	-x	-	+x	+x
Equilibrium	0.6 - x	-	x	x

\[\begin{align*} 1.8 \times 10^{-5} &= \dfrac{x^2}{0.6 - x} \\[4pt] (1.08 \times 10^{-5}) - (1.8 \times 10^{-5})x - x^2 &= 0 \\[4pt] x &= \dfrac{1.8 \times 10^{-5} \pm \sqrt{(1.8 \times 10^{-5})^2 - 4(-1)(1.08 \times 10^{-5})}}{2(-1)} \\[4pt] &= \dfrac{1.8 \times 10^{-5} \pm 6.57 \times 10^{-3}}{-2} \end{align*}\]

The two solutions are thus \([\ce{OH^{-}}] = -3.29 \times 10^{-3}\) and \([\ce{OH^{-}}]= 3.28 \times 10^{-3} \; M \; OH^-\). The first one is meaningless, so from the second value we can calculate the pOH and pH:

\[pOH = -\log(3.28 \times 10^{-3}) = 2.5 \nonumber \]

\[pH = 14 - pOH = 14 - 2.5 = 11.5 \nonumber \]

b) At the equivalence point, the number of moles of \(\ce{HCl}\) added is equal to the initial number of moles of NH3, because the analyte is completely neutralized.

\[0.054 \; mol \; HCl \times \dfrac{1 \; L \; HCl}{mol \; HCl} = 0.054 \; L \; HCl, \; or \; 54 \; mL \; HCl \nonumber \]

c) At the midpoint, pOH = pKb.

\(-\log(1.8 \times 10^{-5}) = 4.74 \; pOH\) \(pH = 14 - pOH = 14 - 4.74 = 9.26 \; pH\)

At the midpoint, the number of moles of HCl added equals half the initial number of moles of NH3. In other words, the number of moles of HCl added at the midpoint is half of the number of moles of HCl added by the equivalence point. Hence:

\[\dfrac{1}{2}(0.054 \; L \; HCl \; added \; at \; equivalence \; point) = 0.027 \; moles \; HCl \; at \; midpoint \nonumber \]

\[\text{Volume of acid needed} = 0.027\ mol\ HCl \times \dfrac{1\ L}{1\ mol\ HCl} = 0.027\ L\ HCl = 27\ mL\ HCl \nonumber \]

d) First, find the moles of HCl in 50 mL of HCl.

\[0.05 \; L \; HCl \times \dfrac{mol \; HCl}{L \; HCl} = 0.05 \; mol \; HCl \nonumber \]

\(NH_3\)	\(H_2O\)	\(\rightleftharpoons\)	\(NH_4^+\)	\(OH^-\)
Initial	0.054 mol	-	0	0
Change	-0.050 mol	-	+0.050 mol	+0.050 mol
Equilibrium	0.004 mol	-	0.050 mol	0.050 mol

Because 50 mL of acid have been added, and we started out with 90 mL of analyte, there are a total of 140 mL of analyte solution at this point. Hence, the molarity of NH3 is the following:

\(\dfrac{0.004 \; mol \; NH_3}{0.140 \; L \; solution \; in \; flask}=0.0286 M\)

The molarity of NH4+ is:

\(\dfrac{0.050 \; mol \; NH_4^{+}}{0.140 \; L \; solution \; in \; flask}=0.357 M\)

Now we can use the Henderson-Hasselbalch approximation:

\(pOH=pK_b+log\dfrac{NH_4^{+}}{NH_3}\)

\(pOH=4.74+log\dfrac{0.357}{0.0286}=5.84 \; pOH\)

\(pH=14-pOH=14-4.84=8.16\)

e) To find the pH at the equivalence point, first calculate the molarity of the NH4+ in the flask at this point.

\(\dfrac{0.054 \; mol \; NH_4{^+}}{0.140 \; L \; analyte \; solution}=0.375M \; NH_4{^+}\) \(K_a=\dfrac{K_w}{K_b}=\dfrac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}}=5.56 \times 10^-10\)

\(NH_4^+\)	\(H_2O\)	\(\rightleftharpoons\)	\(NH_3\)	\(H_3O^+\)
Initial	0.375	--	0	0
Change	-x	--	+x	+x
Equilibrium	0.375 -x	--	x	x

\(5.56 \times 10^{-10}=\dfrac{x^2}{0.375+x}\) \(2.09 \times 10^{-10}+(5.56 \times 10^{-10})x-x^2=0\)

We can use the quadratic equation to solve for x:

\(x=\dfrac{-5.56 \times 10^{-10} \pm \sqrt{(5.56 \times 10^{-10})^2-4(-1)(2.09 \times 10^{-10})}}{2(-1)}\) \(=\dfrac{-5.56 \times 10^{-10} \pm 2.89 \times 10^{-5}}{-2} = -1.45 \times 10^{-5}, \; 1.45 \times 10^{-5} \; M \; H_3O^+\) \(pH=-log(1.45 \times 10^{-5})=4.84 \; pH\)

f) First, find the moles of HCl in 60 mL of HCl.

\(0.06 \; L \times \dfrac{mol\; HCl}{L \; HCl}=0.06 \; mol \; OH^-\)

Find the excess amount of HCl, or the amount added after neutralization has occurred. 0.054 moles of HCl reacted with the NH3 to neutralize it.

\(excess \; HCl=0.06-0.054=0.006 \; mol \; HCl\)

Now we need to find the molarity of HCl in the flask at this point. We started out with 90 mL of NH3 analyte in the flask, and added 60 mL. That gives a total of 150 mL, or 0.150 L of solution in the flask.

\(MOLARITY_{HCl \; in \; flask}=\dfrac{0.006 \; mol}{0.150 \; L \; solution}=0.04 \; M \; HCl\)

Because HCl dissociates into H3O+, equate [HCl] to [H3O+]. Now we have the information to determine pH.

\(pH=-log(0.04)=1.4 \; pH\)